Solve -3x^2+6x+9

$$-3 { x }^{ 2 } +6x+9$$

Factor

$$-3\left(x-3\right)\left(x+1\right)$$

Solution Steps

Factor out 3.

$$3\left(-x^{2}+2x+3\right)$$

Consider -x^{2}+2x+3. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=3 b=-1$$

Rewrite -x^{2}+2x+3 as \left(-x^{2}+3x\right)+\left(-x+3\right).

$$\left(-x^{2}+3x\right)+\left(-x+3\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(x-3\right)-\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x-1\right)$$

Rewrite the complete factored expression.

$$3\left(x-3\right)\left(-x-1\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-3x^{2}+6x+9=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-3\right)\times 9}}{2\left(-3\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-3\right)\times 9}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-6±\sqrt{36+12\times 9}}{2\left(-3\right)}$$

Multiply 12 times 9.

$$x=\frac{-6±\sqrt{36+108}}{2\left(-3\right)}$$

Add 36 to 108.

$$x=\frac{-6±\sqrt{144}}{2\left(-3\right)}$$

Take the square root of 144.

$$x=\frac{-6±12}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-6±12}{-6}$$

Now solve the equation x=\frac{-6±12}{-6} when ± is plus. Add -6 to 12.

$$x=\frac{6}{-6}$$

Divide 6 by -6.

$$x=-1$$

Now solve the equation x=\frac{-6±12}{-6} when ± is minus. Subtract 12 from -6.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and 3 for x_{2}.

$$-3x^{2}+6x+9=-3\left(x-\left(-1\right)\right)\left(x-3\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-3x^{2}+6x+9=-3\left(x+1\right)\left(x-3\right)$$

Evaluate

$$-3\left(x-3\right)\left(x+1\right)$$